Kemi Opgave 28-29 32-34 36-37

Se side 56 og 58 i Kemi B bogen.

Indholdsfortegnelse

1 Opgave 28
2 Opgave 29
3 Opgave (30)
4 Opgave (31)
5 Opgave 32
6 Opgave 33
7 Opgave 34
8 Opgave (35)
9 Opgave 36
10 Opgave (37)

Opgave 28

$LaTex: pH_{HNO_3} = -log(0,12) = 0,92\\ pH_{HCL} = -log(2,5\cdot 10^{-3}) = 2,6\\$

Opgave 29

$LaTex: n = 0,02\cdot 0,12 + 0,08\cdot 2,5\cdot 10^{-3}\\ v = 0,08+0,02\\ [H_3O] = \frac{n}{v}\\ pH = -log([H_3O]) = 1,585\\$

Opgave (30)

$LaTex: v_1 = 0,025\\ n = 10^{-2,3}\cdot v_1\\ v_2 = 0,1\\ C_s = \frac{n}{v_2}\\ pH = -log(C_s) = 2,9\\$

Opgave (31)

A: $LaTex: [H_3O^{+}] = 0,1 M\\ pH = -log([H_3O^{+}]) = 1\\$
B: $LaTex: [H_3O^{+}] = 10^{-4} M\\ pH = -log([H_3O^{+}]) = 4\\$

Opgave 32

$LaTex: pH = \frac{1}{2}\cdot (4,76-log(0,1) = 2,88\\$

Opgave 33

$LaTex: pH = \frac{1}{2}\cdot (4,76-log(0,01) = 3,38\\ [H_3O^{+}] = 10^{-pH} = 0,00042 M\\ [CH_3COO^{-}] = [H_3O^{+}] = 0,00042 M\\ [CH_3COOH] = 0,01-[CH_3COO^{-}] = 0,00958 M\\$

Opgave 34

$LaTex: pH = \frac{1}{2}\cdot (4,76-log(0,01) = 3,38\\ [HCL] = [H_3O] = 10^{-pH} = 0,00042 M\\$

Opgave (35)

$LaTex: C_s = \frac{0,048*1000}{60}\\ pK_s = 4,76\\ pH = \frac{1}{2}\cdot (pK_s -log(C_s)) = 2,43\\$

Opgave 36

$LaTex: n = \frac{2,54g}{53,49\frac{mol}{g}} = 0,0475 mol\\ C_s = \frac{n}{0,5 l}\\ pKs = 9,25\\ pH = \frac{1}{2}\cdot (pKs-log(C_s)) = 5,14\\$

Opgave (37)

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Kategorier: Kemi Opgaver